05/27 - Lab 26: Signals with Multiple Frequency Components

Lab 26: Signals with Multiple Frequency Components

In this lab, we used Waveforms to create complex voltage signals which were applied to a simple RC circuit. The first signal composed of multiple sinusoidal waves of different frequencies, and the second signal was a sinusoidal wave with a time-varying frequency. The response of the circuit was measured in order to see how the "shape" of the input is affected. The following circuit was used in this lab:

The values used in the circuit were R = 1 kOhm, and C = 100 nF, where V_in(t) is the complex sinusoidal signal, and V_out is the voltage measured across the "load" resistor. The actual circuit is pictured below:

For the pre-lab, we calculated the magnitude response of the circuit, which is simply the ratio of the amplitude of the input sinusoid to the amplitude of the output sinusoid, in order to determine the zeros and poles of the circuit.

The function used for our first input sinusoid was a function containing multiple sine functions at various frequencies:

V_in(t) = 20[sin(1000πt) + sin(2000πt) + sin(20,000πt)]

The frequency of the overall sinusoid was adjusted throughout the experiment. We used frequencies of 500 Hz, 1000 Hz, and 10000 Hz. Pictured below is the oscilloscope depicting the input and output voltages for each case:

500 Hz

Yellow: Input Voltage  |  Blue: Output Voltage

1000 Hz

Yellow: Input Voltage  |  Blue: Output Voltage

10000 Hz

Yellow: Input Voltage  |  Blue: Output Voltage

The second input function used in this lab was a sinusoidal sweep function that contained a time-varying frequency. We did not need a mathematical equation for this, as Waveforms has a built-in sweep function. The input wave using this built-in feature looks like this:

Pictured below is the Waveforms oscilloscope containing the input voltage, as well as the voltage across the "load" resistor.

In Class Examples

1. For the circuit shown at right, calculate the gain I o (ω)/I i (ω) and its poles and zeros.

2. Given the circuit shown below and i(t)= I cos ωt amps, find the transfer function H(ω) = Vo / I and sketch the frequency response.

3. Given the circuit shown at right and vin = Vin cos ωt volts, find the transfer function H(ω ) = Vout / Vin where Vout is the voltage across the inductor and sketch the frequency response.

05/24 - Lab 25: Apparent Power and Power Factor

Lab 25: Apparent Power and Power Factor

In this lab, we used a basic LR circuit to gain understanding of apparent power and power factor in order to determine the power delivered to a load. Apparent power, average power, and the power factor associated with the circuit were calculated on our white boards, and then measured during the experiment to compare our expectations with the actual values. The circuit used in this experiment is pictured below:

For this experiment, R_T = 10 Ohms, L = 1 mH, and the load resistance R_L will vary throughout the experiment from 10 Ohms to 47 Ohms, and finally 100 Ohms. The input signal V(t) had an amplitude of 1V @ 5 kHz. The circuit we built in the breadboard is pictured below (with R_L = 10 Ohms).

Our first set of calculations (with R_L = 10 Ohms) is pictured on our white board below:

We predicted average power P = 7.2 mW, apparent power S = 13.4 mVA, V_rms = 0.707, 0° V, and I_rms = 19, -57.61° mA. The actual results from measuring values on our circuit are pictured below:

The actual V_rms turned out to be 591 mV, which is slightly off but still fairly close to our predicted result (only off by ~0.1V), and our measured I_rms current turned out to be 18.3 mA which was only ~0.7 mA from our predicted result. The phase shift of the current from the input voltage can be seen as being ~-60°, which correlates with our predictions.

Next, we changed out R_L from the 10 Ohm resistor and replaced it with a 47 Ohm resistor. Our white board calculations are pictured below:

Our calculations yielded an average power P = 6.8 mW, apparent power S = 7.8 mVA, V_rms = .707, 0° V, and I_rms = 11, -28.8° mA. The actual results of our circuit are pictured below:

The actual voltage obtained from measuring the circuit was V_rms = 608 mV (again, ~0.1V off from our prediction), and our I_rms from measurements turned out to be 10.9 mA, which was extremely close to our prediction of 11 mA. The phase shift of the current can be seen as being slightly negative, which makes sense with our prediction of -28.8°.

Finally, we replaced R_L with a 100 Ohm resistor. Our white board calculations are shown below:

Our calculations yielded an average power P = 4.0 mW, apparent power S = 4.2 mVA, V_rms = .707, 0° V, and I_rms = 5.9, -15.9° mA. The actual measurements from our circuit are pictured below:

The actual voltave V_rms turned out to be 645 mV (~ .05V from our prediction), and the actual current I_rms was slightly off from our prediction at 7.1 mA (predicted 5.9 mA). The phase shift of the current from the input voltage can be seen as very small, and negative, which corresponds nicely to the value we predicted of -15.9°.

In Class Examples

1. Calculate the RMS value of the signal shown in below.

2. A relay coil is connected to a 210-V, 50-Hz supply. If it has a resistance of 30 Ω and an inductance of 0.5 H, calculate the apparent power and the power factor.

3. The voltage across a load is v(t) = 60 cos(ωt − 10◦) V and the current through the element in the direction of the voltage drop is i(t) =1.5 cos(ωt + 50◦). Find complex and apparent powers.

4. When connected to a 120-V (rms), 60-Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.